To find the radius of circle B, use the circumference formula (c = πd = 2πr):

2πr = 36π

Divide each side by 2π: r = 18

Now, if circle A has a radius half the length of that of B, A's radius is 18 / 2 = 9.

The area of a circle is πr².  Therefore, for A, it is π*9² = 81π.

We know the triangle inscribed within the circle must be isosceles, as it contains a 90-degree angle and fixed radii. As such, the opposite angles must be equal. Therefore we can use a simplified version of the Pythagorean Theorem, 

a² + a² = c² → 2r² = 16 → r² = 8; r = √8 < 3. (since we know √9 = 3, we know √8 must be less); therefore, Quantity B is greater. 

A radius of 5 means we need a distance of 5 from the center to any points on the circle. We need 5² = (1 – x₂)² + (2 – y₂)². Let's start with the first point, (3,4). (1 – 3)² + (2 – 4)² ≠ 25. Next let's try (4,6). (1 – 4)² + (2 – 6)² = 25, so (4,6) is our answer. The same can be done for the other three points to prove they are incorrect answers, but this is something to do ONLY if you have enough time. 

This question is easy on the whole, though you must not be intimidated by one small fact that we will soon see. Set up your standard circumference equation:

$\displaystyle C = 2\pi r$

The circumference is $\displaystyle 215$ feet, so we can say:

$\displaystyle 215 = 2\pi r$

Solving for $\displaystyle r$, we get:

$\displaystyle r = \frac{215}{2\pi} = \frac{107.5}{\pi}$

Some students may be intimidated by having $\displaystyle \pi$ in the denominator; however, there is no need for such intimidation. This is simply the answer!

Since we know that the area of Square $\displaystyle ABCD$ is $\displaystyle 1156$, we know $\displaystyle 1156 = s^2$, where $\displaystyle s$ is the length of one of its sides. From this, we can solve for $\displaystyle s$ by taking the square root of both sides. You will have to do this by estimating upward. Therefore, you know that $\displaystyle 30^2$ is $\displaystyle 900$. By careful guessing, you can quickly see that $\displaystyle 34^2$ is $\displaystyle 1156$. From this, you know that the diameter of your circle must be half of $\displaystyle 34$, or $\displaystyle 17$ (because it is circumscribed). Therefore, you can draw:

The diameter is

$\displaystyle 2\cdot \text{Radius }$

To solve for the largest diameter multiply each side by 2.  

The resulting formula for diamenter is

 $\displaystyle \text{Diameter }= \frac{S}{\sqrt{3}}$.  

Substitute in 5  for S and solve. Diameter = $\displaystyle \frac{5}{\sqrt{3}}$ = 2.89 cm 

Consider each quantity separately.

Quantity A

Recall that the area of a circle is defined as:

$\displaystyle A=\pi r^2$

We know that the area is $\displaystyle 81\pi$. Therefore,

$\displaystyle 81\pi=\pi r^2$

Divide both sides by $\displaystyle \pi$:

$\displaystyle r^2=81$

Therefore, $\displaystyle r=9$. Since $\displaystyle d=2r$, we know:

$\displaystyle d=2*9=18$

Quantity B

This is very easy. Recall that:

$\displaystyle C=\pi d$

Therefore, if $\displaystyle C=30\pi$, $\displaystyle d=30$. Therefore, Quantity B is larger.

Consider each quantity separately.

Quantity A

Recall that the area of a circle is defined as:

$\displaystyle A=\pi r^2$

We know that the area is $\displaystyle 109\pi$. Therefore,

$\displaystyle 109\pi=\pi r^2$

Divide both sides by $\displaystyle \pi$:

$\displaystyle r^2=109$

Therefore, $\displaystyle r=\sqrt{109}$.  Since $\displaystyle d=2r$, we know:

$\displaystyle d=2\sqrt{109}$

Quantity B

This is very easy.  Recall that:

$\displaystyle C=\pi d$

Therefore, if $\displaystyle C=22\pi$, $\displaystyle d=22$.  

Now, since your calculator will not have a square root button on it, we need to estimate for Quantity A. We know that $\displaystyle \sqrt{121}$ is $\displaystyle 11$. Therefore, $\displaystyle \sqrt{109}< 11$.  This means that $\displaystyle 2 * \sqrt{109} < 22$. Therefore, Quantity B is larger.

From the ratio given $\displaystyle 1:2:3$, it may be easier to write it such that the terms sum up to $\displaystyle 1$. This can be taken by dividing each term by the sum of the terms:

$\displaystyle \frac{1}{6}:\frac{2}{6}:\frac{3}{6}$ 

or

$\displaystyle \frac{1}{6}:\frac{1}{3}:\frac{1}{2}$

The largest sector thus has an area equal to

$\displaystyle \frac{1}{2}30\pi=15\pi$

To find the area of the circle, it is important to know either its diameter or radius. For the geometry described in this problem, this is the same as the diagonal of the rectangle.

However, to find the diagonal of the rectangle, the sides must first be known. They can be found, since the perimeter and area are given:

$\displaystyle lw=360in^2$

$\displaystyle 2l+2w=98in$

This system of equation can be solved by substitution:

$\displaystyle 2l=98in-2w$

$\displaystyle l=49in-w$

Followed by:

$\displaystyle w(49in-w)=360in^2$

$\displaystyle w^2-49in(w)+360in^2=0$

$\displaystyle (w-9in)(w-40in)=0$

Note that this gives two possible values for $\displaystyle w$, $\displaystyle 9in$ or $\displaystyle 40in$, though the one selected is irrelevant; the other value will be the value for $\displaystyle l$.

Knowing these two values, the diagonal can be found; it is the hypotenuse of a right triangle formed by these two lengths:

$\displaystyle d^2=9^2+40^2$

$\displaystyle d=41$

Since the diagonal is also the diameter, the area of the circle is given by:

$\displaystyle \pi \left(\frac{d}{2}\right)^2=\pi \frac{1681^}{4}=420.25\pi$